Limits of Sums and Difference of Complex-Valued Functions

Recall that if $(S, d_S)$ and $(T, d_T)$ are both metric spaces, $A \subseteq S$, $p \in S$ is an accumulation point of $A$, and $f : A \to T$ then we say that $\lim_{x \to p} f(x) = b$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in D(f) \setminus \{ p \} = A \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $d_T(f(x), b) < \epsilon$.

We have already looked at functions $f : A \to \mathbb{R}$ where $A \subseteq \mathbb{R}$, and we will now look at a general class of functions from a metric space $(S, d_S)$ into the metric space on complex numbers $(\mathbb{C}, d)$ where $d$ is the usual metric defined for all $x, y \in \mathbb{C}$ by:

(1)

\begin{align} \quad d(x, y) = \mid x - y \mid \end{align}

Many of the theorems for the limits of functions from a subset of the real numbers to the real numbers also hold for limits of functions from an arbitrary metric space into the complex numbers with the usual metric.

Theorem 1: Let $(S, d_S)$ and $(\mathbb{C}, d)$ be metric spaces where $d$ is the usual metric defined for all $x, y \in \mathbb{C}$ by $d(x, y) = \mid x - y \mid$. Also let $A \subseteq S$, $p \in S$ be an accumulation point of $A$, and $f, g : A \to \mathbb{C}$. Suppose that $\lim_{x \to p} f(x) = a$ and $lim_{x \to p} g(x) = b$. Then $\lim_{x \to p} [f(x) + g(x)] = a + b$.

• Proof: Suppose that $\lim_{x \to p} f(x) = a$ and $\lim_{x \to p} g(x) = b$. We want to prove that $\lim_{x \to p} [f(x) + g(x)] = a + b$, i.e., show that for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $d_S(x, p) < \delta$ then $d(f(x) + g(x), a + b) < \epsilon$.

• Since $\lim_{x \to p} f(x) = A$ then for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a $\delta_1 > 0$ such that if $x \in D(f) = A$ and $d_S(x, p) < \delta_1$ then $d(f(x), a) < \epsilon_1 = \frac{\epsilon}{2}$, $(*)$.

• Similarly, since $\lim_{x \to p} g(x) = A$ then for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists a $\delta_2 > 0$ such that if $x \in D(g) = A$ and $d_S(x, p) < \delta_2$ then $d(g(x), b) < \epsilon_2 = \frac{\epsilon}{2}$, $(**)$.

• Let $\delta = \min \{ \delta_1, \delta_2 \}$. Then if $x \in D(f) = A$ and $d_S(x, p) < \delta$ then $d_S(x, p) < \delta_1$ AND $d_S(x, p) < \delta_2$, so both $(*)$ and $(**)$ will hold, and so:

(2)

\begin{align} \quad \: d(f(x) + g(x), a + b) = \mid [f(x) + g(x)] - [a + b] \mid = \mid [f(x) - a] + [g(x) - b] \mid \leq \mid f(x) - a \mid + \mid g(x) - b \mid &= d(f(x), a) + d(g(x), b) \\ & < \epsilon_1 + \epsilon_2 \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ & <\epsilon \end{align}

• Therefore $\lim_{x \to p} [f(x) + g(x)] = a + b$. $\blacksquare$

Theorem 2: Let $(S, d_S)$ and $(\mathbb{C}, d)$ be metric spaces where $d$ is the usual metric defined for all $x, y \in \mathbb{C}$ by $d(x, y) = \mid x - y \mid$. Also let $A \subseteq S$, $p \in S$ be an accumulation point of $A$, and $f, g : A \to \mathbb{C}$. Suppose that $\lim_{x \to p} f(x) = a$ and $lim_{x \to p} g(x) = b$. Then $\lim_{x \to p} [f(x) - g(x)] = a - b$.

• Proof: Suppose that $\lim_{x \to p} f(x) = a$ and $\lim_{x \to p} g(x) = b$. We want to prove that $\lim_{x \to p} [f(x) - g(x)] = a - b$, i.e., show that for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $d_S(x, p) < \delta$ then $d(f(x) - g(x), a - b) < \epsilon$.

• Since $\lim_{x \to p} f(x) = A$ then for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a $\delta_1 > 0$ such that if $x \in D(f) = A$ and $d_S(x, p) < \delta_1$ then $d(f(x), a) < \epsilon_1 = \frac{\epsilon}{2}$, $(*)$.

• Similarly, since $\lim_{x \to p} g(x) = A$ then for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists a $\delta_2 > 0$ such that if $x \in D(g) = A$ and $d_S(x, p) < \delta_2$ then $d(g(x), b) < \epsilon_2 = \frac{\epsilon}{2}$, $(**)$.

• Let $\delta = \min \{ \delta_1, \delta_2 \}$. Then if $x \in D(f) = A$ and $d_S(x, p) < \delta$ then $d_S(x, p) < \delta_1$ AND $d_S(x, p) < \delta_2$, so both $(*)$ and $(**)$ will hold, and so:

(3)

\begin{align} \quad \: d(f(x) - g(x), a - b) = \mid [f(x) - g(x)] - [a - b] \mid = \mid [f(x) - a] + [b - g(x)] \mid \leq \mid f(x) - a \mid + \mid g(x) - b \mid &= d(f(x), a) + d(g(x), b) \\ & < \epsilon_1 + \epsilon_2 \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ & < \epsilon \end{align}

• Therefore $\lim_{x \to p} [f(x) + g(x)] = a + b$. $\blacksquare$